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(B) The ink droplet enters the region between the plates with horizontal velocity $v$. The electric field $E$ is directed downwards. Since the droplet

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$\frac{mv^2d}{EL^2}$

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(B) The ink droplet enters the region between the plates with horizontal velocity $v$. The electric field $E$ is directed downwards. Since the droplet has a negative charge $q$,it experiences an upward electric force $F = qE$.The acceleration of the droplet in the vertical direction is $a = \frac{F}{m} = \frac{qE}{m}$.The time taken to travel the length $L$ of the plates is $t = \frac{L}{v}$.For the droplet not to hit the upper plate,the vertical displacement $y$ must be less than or equal to half the distance between the plates,i.e.,$y \le \frac{d}{2}$.Using the equation of motion $y = \frac{1}{2}at^2$,we have:$\frac{d}{2} = \frac{1}{2} \left( \frac{qE}{m} \right) \left( \frac{L}{v} \right)^2$Solving for $q$:$d = \frac{qE L^2}{m v^2}$$q = \frac{m v^2 d}{E L^2}$Thus,the maximum charge is $\frac{m v^2 d}{E L^2}$.

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